Use linear approximation, i. e. the tangent line, to approximate 1/0.201 as follows: let f (x) = 1/x and find the equation of the tangent line to f(x) at a "nice" point near 0.201 . then use this to approximate 1/0.201.

Write the equation of a line that is perpendicular to the given line and that passes through the given point. –3x – 6y = 17; (6, 3) y = x – 9 y = 2x – 9 y = –2x – 9 y = x + 0 3. is the relationship shown by the data linear? if so, model the data with an equation. x y 1 5 5 10 9 15 13 20 the relationship is linear; y – 5 = (x – 1). the relationship is not linear. the relationship is linear; y – 5 = (x – 1). the relationship is linear; y – 1 = (x – 5). write an equation in point-slope form for the line through the given point with the given slope. (–10, –1); m = y + 10 = (x + 1) y – 1 = (x – 10) y – 1 = (x + 10) y + 1 = (x + 10) 5. write an equation for each translation of . 6.5 units up y + 6.5 = | x | y = | 6.5 x | y = | x | + 6.5 y = | x | – 6.5 6. write an equation for each translation of . 5.5 units right y = | x | + 5.5 y = | x – 5.5 | y = | x | – 5.5 y = | x + 5.5 | 7. which equation translates y = | x | by 8 units to the left? y = | x | – 8 y = | x | + 8 y = | x – 8| y = | x + 8|